Calculus

Determine the limit

lim x->0 1-cos^2(3x)/x^2

asked by Seeina
  1. lim x->0 1-cos^2(3x)/x^2
    Since it is a situation of 0/0, we can use l'Hôpital's rule by differentiating both numerator and denominator to get:
    lim x->0 1-cos^2(3x)/x^2
    =lim x->0 6sin(3x)/2x
    Since it is still a situation of 0/0, we can apply l'Hôpital's rule again:
    =lim x->0 18cos(3x)/2
    = 18*1/2
    = 9

    If you have learned about Taylor series or MacLauin series, you can expand the numerator into a series, and evaluate accordingly:

    lim x->0 1-cos^2(3x)/x^2
    =lim x->0 1-(1-(3x)^2/2!+(3x)^4/4!-...)^2/x^2
    =lim x->0 1-(1-9x^2+27x^4-...)/x^2
    =lim x->0 (9x^2-27x^4)/x^2
    =lim x->0 (9-27x^2+...)
    =9

    posted by MathMate

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